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Optimization Problems

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So Shreya,
Once you set this up as a math problem you're halfway to the answer. Let "x" be the distance cut off for the piece you're going to bend into a square. Then the piece bent into a circle is what length?
Now write the an expression for the sum of the areas covered by the two finished forms. Since "x" is the total for the square perimeter, how long is each side of the square? Then what is the area of the square?
Next, for the circle, you have the length of the circumference. What are the area of a circle and its circumference  A =(pi/4) d^2 ; C =pi d? Get an expression for d from the length of the circumference, and plug it into the expression for area of the circle. You should now have (after combining some factors): areatotal = (x/4)^2 + (50-(x/4))^2/pi . When nature gives you such a nice equation for (x/4), don't go solving for x! Instead, substitute another "dummy" variable, call it y, in place of (x/4) throughout. You then have: area = y^2 = (50-y)^2/pi.
Expand this: area = y^2 + 2500/pi -100y/pi +y^2/pi
Take the first derivative with respect to y: if there is a minimum to the area function, the first derivative will be zero there:
dA/dy = 2y -100/pi +2y/pi
Solve for this = 0:
y(1+1/pi)-50/pi = 0
y= (50/pi)/((pi+1)/pi)
y=50/(pi+1)
Now substitute back to get x (the distance you needed to cut at) x = 4y.
But, does this give you a minimum area or a maximum area? To test this easily, take and examine the second derivative of the area with respect to y (take the next derivative of the dA/dy equation):
d''A/dy^2 = 2 + 2/pi. This is positive; hence, the area function is concave upwards for all values of y, and if there is a point of zero slope between the two extremes (0 and 50 cm for y) it must be a minimum.
Pretty simple, and it gives you a fixed ratio you'd need to cut any length wire along to do the same thing, namely, 1/(1+pi) of the way along it.
By the way, don't forget to bend the smaller piece into the square, and the larger into the circle....